Let #ABCD# be a trapezoid with lower base #AD# and upper base #BC#.
#M# is a midpoint of left leg #AB# and #N# is a midpoint of right leg #CD#.
Connect vertex #B# with midpoint #N# of opposite leg #CD# and extend it beyond point #N# to intersect with continuation of lower base #AD# at point #X#.
Consider two triangles #Delta BCN# and #Delta NDX#. They are congruent by angle-side-angle theorem because
(a) angles ∠#BNC# and ∠#DNX# are vertical,
(b) segments #CN# and #ND# are congruent (since point #N# is a midpoint of segment #CD#),
(c) angles ∠#BCN# and ∠#NDX# are alternate interior angles with parallel lines #BC# and #DX# and transversal #CD#.
Therefore, segments #BC# and #DX# are congruent, as well as segments #BN# and #NX#, which implies that #N# is a midpoint of segment #BX#.
Now consider triangle #Delta ABX#. Since #M# is a midpoint of leg #AB# by a premise of this theorem and #N# is a midpoint of segment #BX#, as was just proven, segment #MN# is a mid-segment of triangle #Delta ABX# and, therefore, is parallel to its base #AX# and equal to its half.
But #AX# is a sum of lower base #AD# and segment #DX#, which is congruent to upper base #BC#. Therefore, #MN# is equal to half of sum of two bases #AD# and #BC#.
End of proof.
The lecture dedicated to this and other properties of quadrilaterals as well as many other topics are addressed by a course of advanced math for high school students at Unizor.
